Subgroups of a group of even order

My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

Thanks!
-Dan

I assume that you mean no non-trivial proper subgroups. The answer is yes but in the spirit of your question...no. If is a finite group and where then a simple application of Cauchy's theorem says that must have an element of order . More generally, Sylow's first theorem implies that if is a finite group and with prime then must contain a subgroup of order . From this it's easy to see that the only groups you mention have order or where is prime.

I assume that you mean no non-trivial proper subgroups. The answer is yes but in the spirit of your question...no. If is a finite group and where then a simple application of Cauchy's theorem says that must have an element of order . More generally, Sylow's first theorem implies that if is a finite group and with prime then must contain a subgroup of order . From this it's easy to see that the only groups you mention have order or where is prime.

Thanks. "Cauchy" isn't until the next chapter, but I can follow the logic easily enough. Thanks!